2391. Minimum Amount of Time to Collect Garbage LeetCode Solution

Last updated: February 27, 2025

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Problem Statement
Complexity Analysis
Minimum Amount of Time to Collect Garbage solution in C++
Minimum Amount of Time to Collect Garbage solution in Java
Minimum Amount of Time to Collect Garbage solution in Python
Additional Resources

2391. Minimum Amount of Time to Collect Garbage LeetCode Solution image

Problem Statement of Minimum Amount of Time to Collect Garbage

You are given a 0-indexed array of strings garbage where garbage[i] represents the assortment of garbage at the ith house. garbage[i] consists only of the characters ‘M’, ‘P’ and ‘G’ representing one unit of metal, paper and glass garbage respectively. Picking up one unit of any type of garbage takes 1 minute.
You are also given a 0-indexed integer array travel where travel[i] is the number of minutes needed to go from house i to house i + 1.
There are three garbage trucks in the city, each responsible for picking up one type of garbage. Each garbage truck starts at house 0 and must visit each house in order; however, they do not need to visit every house.
Only one garbage truck may be used at any given moment. While one truck is driving or picking up garbage, the other two trucks cannot do anything.
Return the minimum number of minutes needed to pick up all the garbage.

Example 1:

Input: garbage = [“G”,”P”,”GP”,”GG”], travel = [2,4,3]
Output: 21
Explanation:
The paper garbage truck:
1. Travels from house 0 to house 1
2. Collects the paper garbage at house 1
3. Travels from house 1 to house 2
4. Collects the paper garbage at house 2
Altogether, it takes 8 minutes to pick up all the paper garbage.
The glass garbage truck:
1. Collects the glass garbage at house 0
2. Travels from house 0 to house 1
3. Travels from house 1 to house 2
4. Collects the glass garbage at house 2
5. Travels from house 2 to house 3
6. Collects the glass garbage at house 3
Altogether, it takes 13 minutes to pick up all the glass garbage.
Since there is no metal garbage, we do not need to consider the metal garbage truck.
Therefore, it takes a total of 8 + 13 = 21 minutes to collect all the garbage.

Example 2:

Input: garbage = [“MMM”,”PGM”,”GP”], travel = [3,10]
Output: 37
Explanation:
The metal garbage truck takes 7 minutes to pick up all the metal garbage.
The paper garbage truck takes 15 minutes to pick up all the paper garbage.
The glass garbage truck takes 15 minutes to pick up all the glass garbage.
It takes a total of 7 + 15 + 15 = 37 minutes to collect all the garbage.

Constraints:

2 <= garbage.length <= 105
garbage[i] consists of only the letters 'M', 'P', and 'G'.
1 <= garbage[i].length <= 10
travel.length == garbage.length – 1
1 <= travel[i] <= 100

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

2391. Minimum Amount of Time to Collect Garbage LeetCode Solution in C++

class Solution {
 public:
  int garbageCollection(vector<string>& garbage, vector<int>& travel) {
    vector<int> prefix(travel.size());
    partial_sum(travel.begin(), travel.end(), prefix.begin());
    const int timeM = getTime(garbage, prefix, 'M');
    const int timeP = getTime(garbage, prefix, 'P');
    const int timeG = getTime(garbage, prefix, 'G');
    return timeM + timeP + timeG;
  }

 private:
  int getTime(const vector<string>& garbage, const vector<int>& prefix,
              char c) {
    int characterCount = 0;
    int lastIndex = -1;
    for (int i = 0; i < garbage.size(); ++i) {
      const string& s = garbage[i];
      if (ranges::any_of(s, [c](const char g) { return g == c; }))
        lastIndex = i;
      characterCount += std::ranges::count(s, c);
    }
    return characterCount + (lastIndex <= 0 ? 0 : prefix[lastIndex - 1]);
  }
};
/* code provided by PROGIEZ */

2391. Minimum Amount of Time to Collect Garbage LeetCode Solution in Java

class Solution {
  public int garbageCollection(String[] garbage, int[] travel) {
    int[] prefix = new int[travel.length];
    prefix[0] = travel[0];
    for (int i = 1; i < prefix.length; ++i)
      prefix[i] += prefix[i - 1] + travel[i];
    final int timeM = getTime(garbage, prefix, 'M');
    final int timeP = getTime(garbage, prefix, 'P');
    final int timeG = getTime(garbage, prefix, 'G');
    return timeM + timeP + timeG;
  }

  private int getTime(String[] garbage, int[] prefix, char c) {
    int characterCount = 0;
    int lastIndex = -1;
    for (int i = 0; i < garbage.length; ++i) {
      final String s = garbage[i];
      if (s.chars().anyMatch(g -> g == c))
        lastIndex = i;
      characterCount += (int) s.chars().filter(g -> g == c).count();
    }
    return characterCount + (lastIndex <= 0 ? 0 : prefix[lastIndex - 1]);
  }
}
// code provided by PROGIEZ

2391. Minimum Amount of Time to Collect Garbage LeetCode Solution in Python

class Solution:
  def garbageCollection(self, garbage: list[str], travel: list[int]) -> int:
    prefix = list(itertools.accumulate(travel))

    def getTime(c: str) -> int:
      characterCount = 0
      lastIndex = -1
      for i, s in enumerate(garbage):
        if any(g == c for g in s):
          lastIndex = i
        characterCount += s.count(c)
      return characterCount + (0 if lastIndex <= 0 else prefix[lastIndex - 1])

    return getTime('M') + getTime('P') + getTime('G')
 # code by PROGIEZ

Additional Resources

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